multiplicative abelian linearly ordered group on $Bbb R-{0}$
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It seems that $Bbb Rsetminus{0}$ does not create an alo-group (abelian linearly ordered group) with the "normal" order on $Bbb R$.
Is it possible to define any other order (including a trivial order) so we can have alo-group on $Bbb R-{0}$?
Answer to this question may lead to a joint publication.
group-theory order-theory
edited Nov 17 at 3:25
Tianlalu
2,8811935
asked Nov 17 at 3:05
Jan Tamariani
2
|
show 1 more comment
up vote
-2
down vote
favorite
It seems that $Bbb Rsetminus{0}$ does not create an alo-group (abelian linearly ordered group) with the "normal" order on $Bbb R$.
Is it possible to define any other order (including a trivial order) so we can have alo-group on $Bbb R-{0}$?
Answer to this question may lead to a joint publication.
group-theory order-theory
edited Nov 17 at 3:25
Tianlalu
2,8811935
asked Nov 17 at 3:05
Jan Tamariani
2
There is a sentence in en.wikipedia.org/wiki/Linearly_ordered_group which says "F. W. Levi showed that an abelian group admits a linear order if and only if it is torsion free". Does this answer your question?
– Joppy
Nov 17 at 3:35
this is an excellent hint! Thank you. Unfortunately, this is the answer. However, it looks like my question is not as trivial as it looks like. On the other hand, real numbers w/o 0 create a multiplicative torsion free abelian group. So, by F.W. Levi, there must be a linear order hence it is also abelian linearly ordered group! I wonder if anyone of you, wise people, can confirm it. I specialize in computers and feel like an intruder here :)
– Jan Tamariani
Nov 18 at 11:02
the hint has been excellent but it does not solve the problem. The problem is with the condition: if x (?) y than zx (?) zy where x, y, z are group elements and (?) is the linear order. For the "normal" order "<" for real numbers and z=-1, < changes to > hence negative numbers create a problem for the regular "order". I wonder if = can be used as (?) in some deviated way. How about neq ? It looks like a partial order to me.
– Jan Tamariani
Dec 1 at 0:40
The group $mathbb{R} - {0}$ with multiplication has torsion, since the element $-1$ squares to $1$.
– Joppy
Dec 1 at 2:00
Oh great mathematical brothers and sisters :) I allow myself for such opening since I have "applied math" as a part of my MSc degree title (or description). Later on, I have added to it PhD in what you be descibed as 0-1 :) which are computers. They totally destroyed my math brains :) I wonder if it is "legal" to reveal my true name here and if it kills you to write a few human words. My understanding is
– Jan Tamariani
yesterday
|
show 1 more comment
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
It seems that $Bbb Rsetminus{0}$ does not create an alo-group (abelian linearly ordered group) with the "normal" order on $Bbb R$.
Is it possible to define any other order (including a trivial order) so we can have alo-group on $Bbb R-{0}$?
Answer to this question may lead to a joint publication.
group-theory order-theory
edited Nov 17 at 3:25
Tianlalu
2,8811935
asked Nov 17 at 3:05
Jan Tamariani
2
It seems that $Bbb Rsetminus{0}$ does not create an alo-group (abelian linearly ordered group) with the "normal" order on $Bbb R$.
Is it possible to define any other order (including a trivial order) so we can have alo-group on $Bbb R-{0}$?
Answer to this question may lead to a joint publication.
group-theory order-theory
group-theory order-theory
edited Nov 17 at 3:25
Tianlalu
2,8811935
asked Nov 17 at 3:05
Jan Tamariani
2
edited Nov 17 at 3:25
Tianlalu
2,8811935
asked Nov 17 at 3:05
Jan Tamariani
2
edited Nov 17 at 3:25
Tianlalu
2,8811935
edited Nov 17 at 3:25
Tianlalu
2,8811935
edited Nov 17 at 3:25
Tianlalu
2,8811935
2,8811935
asked Nov 17 at 3:05
Jan Tamariani
2
asked Nov 17 at 3:05
Jan Tamariani
2
asked Nov 17 at 3:05
Jan Tamariani
2
2
There is a sentence in en.wikipedia.org/wiki/Linearly_ordered_group which says "F. W. Levi showed that an abelian group admits a linear order if and only if it is torsion free". Does this answer your question?
– Joppy
Nov 17 at 3:35
this is an excellent hint! Thank you. Unfortunately, this is the answer. However, it looks like my question is not as trivial as it looks like. On the other hand, real numbers w/o 0 create a multiplicative torsion free abelian group. So, by F.W. Levi, there must be a linear order hence it is also abelian linearly ordered group! I wonder if anyone of you, wise people, can confirm it. I specialize in computers and feel like an intruder here :)
– Jan Tamariani
Nov 18 at 11:02
the hint has been excellent but it does not solve the problem. The problem is with the condition: if x (?) y than zx (?) zy where x, y, z are group elements and (?) is the linear order. For the "normal" order "<" for real numbers and z=-1, < changes to > hence negative numbers create a problem for the regular "order". I wonder if = can be used as (?) in some deviated way. How about neq ? It looks like a partial order to me.
– Jan Tamariani
Dec 1 at 0:40
The group $mathbb{R} - {0}$ with multiplication has torsion, since the element $-1$ squares to $1$.
– Joppy
Dec 1 at 2:00
Oh great mathematical brothers and sisters :) I allow myself for such opening since I have "applied math" as a part of my MSc degree title (or description). Later on, I have added to it PhD in what you be descibed as 0-1 :) which are computers. They totally destroyed my math brains :) I wonder if it is "legal" to reveal my true name here and if it kills you to write a few human words. My understanding is
– Jan Tamariani
yesterday
|
show 1 more comment
There is a sentence in en.wikipedia.org/wiki/Linearly_ordered_group which says "F. W. Levi showed that an abelian group admits a linear order if and only if it is torsion free". Does this answer your question?
– Joppy
Nov 17 at 3:35
this is an excellent hint! Thank you. Unfortunately, this is the answer. However, it looks like my question is not as trivial as it looks like. On the other hand, real numbers w/o 0 create a multiplicative torsion free abelian group. So, by F.W. Levi, there must be a linear order hence it is also abelian linearly ordered group! I wonder if anyone of you, wise people, can confirm it. I specialize in computers and feel like an intruder here :)
– Jan Tamariani
Nov 18 at 11:02
the hint has been excellent but it does not solve the problem. The problem is with the condition: if x (?) y than zx (?) zy where x, y, z are group elements and (?) is the linear order. For the "normal" order "<" for real numbers and z=-1, < changes to > hence negative numbers create a problem for the regular "order". I wonder if = can be used as (?) in some deviated way. How about neq ? It looks like a partial order to me.
– Jan Tamariani
Dec 1 at 0:40
The group $mathbb{R} - {0}$ with multiplication has torsion, since the element $-1$ squares to $1$.
– Joppy
Dec 1 at 2:00
Oh great mathematical brothers and sisters :) I allow myself for such opening since I have "applied math" as a part of my MSc degree title (or description). Later on, I have added to it PhD in what you be descibed as 0-1 :) which are computers. They totally destroyed my math brains :) I wonder if it is "legal" to reveal my true name here and if it kills you to write a few human words. My understanding is
– Jan Tamariani
yesterday
There is a sentence in en.wikipedia.org/wiki/Linearly_ordered_group which says "F. W. Levi showed that an abelian group admits a linear order if and only if it is torsion free". Does this answer your question?
– Joppy
Nov 17 at 3:35
There is a sentence in en.wikipedia.org/wiki/Linearly_ordered_group which says "F. W. Levi showed that an abelian group admits a linear order if and only if it is torsion free". Does this answer your question?
– Joppy
Nov 17 at 3:35
this is an excellent hint! Thank you. Unfortunately, this is the answer. However, it looks like my question is not as trivial as it looks like. On the other hand, real numbers w/o 0 create a multiplicative torsion free abelian group. So, by F.W. Levi, there must be a linear order hence it is also abelian linearly ordered group! I wonder if anyone of you, wise people, can confirm it. I specialize in computers and feel like an intruder here :)
– Jan Tamariani
Nov 18 at 11:02
this is an excellent hint! Thank you. Unfortunately, this is the answer. However, it looks like my question is not as trivial as it looks like. On the other hand, real numbers w/o 0 create a multiplicative torsion free abelian group. So, by F.W. Levi, there must be a linear order hence it is also abelian linearly ordered group! I wonder if anyone of you, wise people, can confirm it. I specialize in computers and feel like an intruder here :)
– Jan Tamariani
Nov 18 at 11:02
the hint has been excellent but it does not solve the problem. The problem is with the condition: if x (?) y than zx (?) zy where x, y, z are group elements and (?) is the linear order. For the "normal" order "<" for real numbers and z=-1, < changes to > hence negative numbers create a problem for the regular "order". I wonder if = can be used as (?) in some deviated way. How about neq ? It looks like a partial order to me.
– Jan Tamariani
Dec 1 at 0:40
the hint has been excellent but it does not solve the problem. The problem is with the condition: if x (?) y than zx (?) zy where x, y, z are group elements and (?) is the linear order. For the "normal" order "<" for real numbers and z=-1, < changes to > hence negative numbers create a problem for the regular "order". I wonder if = can be used as (?) in some deviated way. How about neq ? It looks like a partial order to me.
– Jan Tamariani
Dec 1 at 0:40
The group $mathbb{R} - {0}$ with multiplication has torsion, since the element $-1$ squares to $1$.
– Joppy
Dec 1 at 2:00
The group $mathbb{R} - {0}$ with multiplication has torsion, since the element $-1$ squares to $1$.
– Joppy
Dec 1 at 2:00
Oh great mathematical brothers and sisters :) I allow myself for such opening since I have "applied math" as a part of my MSc degree title (or description). Later on, I have added to it PhD in what you be descibed as 0-1 :) which are computers. They totally destroyed my math brains :) I wonder if it is "legal" to reveal my true name here and if it kills you to write a few human words. My understanding is
– Jan Tamariani
yesterday
Oh great mathematical brothers and sisters :) I allow myself for such opening since I have "applied math" as a part of my MSc degree title (or description). Later on, I have added to it PhD in what you be descibed as 0-1 :) which are computers. They totally destroyed my math brains :) I wonder if it is "legal" to reveal my true name here and if it kills you to write a few human words. My understanding is
– Jan Tamariani
yesterday
|
show 1 more comment
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There is a sentence in en.wikipedia.org/wiki/Linearly_ordered_group which says "F. W. Levi showed that an abelian group admits a linear order if and only if it is torsion free". Does this answer your question?
– Joppy
Nov 17 at 3:35
this is an excellent hint! Thank you. Unfortunately, this is the answer. However, it looks like my question is not as trivial as it looks like. On the other hand, real numbers w/o 0 create a multiplicative torsion free abelian group. So, by F.W. Levi, there must be a linear order hence it is also abelian linearly ordered group! I wonder if anyone of you, wise people, can confirm it. I specialize in computers and feel like an intruder here :)
– Jan Tamariani
Nov 18 at 11:02
the hint has been excellent but it does not solve the problem. The problem is with the condition: if x (?) y than zx (?) zy where x, y, z are group elements and (?) is the linear order. For the "normal" order "<" for real numbers and z=-1, < changes to > hence negative numbers create a problem for the regular "order". I wonder if = can be used as (?) in some deviated way. How about neq ? It looks like a partial order to me.
– Jan Tamariani
Dec 1 at 0:40
The group $mathbb{R} - {0}$ with multiplication has torsion, since the element $-1$ squares to $1$.
– Joppy
Dec 1 at 2:00
Oh great mathematical brothers and sisters :) I allow myself for such opening since I have "applied math" as a part of my MSc degree title (or description). Later on, I have added to it PhD in what you be descibed as 0-1 :) which are computers. They totally destroyed my math brains :) I wonder if it is "legal" to reveal my true name here and if it kills you to write a few human words. My understanding is
– Jan Tamariani
yesterday